Number Sequence
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Problem Description
A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
Output
For each test case, print the value of f(n) on a single line.
Sample Input
1 1 3
1 2 10
0 0 0
Sample Output
2
5
仔细看一下数据范围,n的取值达到一亿。由于数据范围很大,因此这题不适合暴力。
那么这道类似斐波那契数列的题目该如何搞呢?很明显,f(n-1)和f(n-2)的取值只可能有0, 1, 2, 3, 4, 5, 6这七种情况。因此f(n-1)+f(n-2)的组合一共有7*7种情况,这说明什么呢?说明数列(以f3作为第一个数)出现循环最多在第50个数(最坏的情况,可由鸽巢原理得出)。
#include<cstdio> int main() { int a,b,n; while(scanf("%d%d%d",&a,&b,&n)!=EOF) { int ans[55]={0}; int f1,f2,i; if(a==0&&b==0&&n==0) break; f1=f2=1; for(i=0;;i++) { ans[i]=(a*f2+b*f1)%7; f1=f2; f2=ans[i]; if(i>=3&&ans[i-1]==ans[0]&&ans[i]==ans[1]) break; } if(n==1||n==2) printf("1\n"); else printf("%d\n",ans[(n-3)%(i-1)]); } return 0; }
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